The resistance $$R$$ of a wire is given by the formula:

$$ R = \rho \frac{l}{A}, $$

where:

• $$\rho$$ is the resistivity of the material,
• $$l$$ is the length of the wire,
• $$A$$ is the cross-sectional area of the wire.

If we have a cylindrical wire, the cross-sectional area can be expressed as $$A = \pi r^2$$, where $$r$$ is the radius of the cylinder. Therefore, the resistance of the original wire can be written as:

$$ R = \rho \frac{l}{\pi r^2}. $$

When the wire is stretched such that its radius becomes $$r / 2$$, its volume would remain constant, given that the volume of a cylinder is $$V = A \cdot l = \pi r^2 \cdot l$$. Assuming the volume before and after the stretching is the same, and since the area is now a quarter of the original (because when the radius is halved, the area, which is proportional to the square of the radius, is reduced to a quarter), the length must have increased to four times the original to preserve the volume. That is,

$$ l' = 4l, $$

and the new area,

$$ A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}. $$

Therefore, the new resistance, $$R'$$, of the wire can be calculated using the original formula for resistance:

$$ R' = \rho \frac{l'}{A'} = \rho \frac{4l}{\frac{\pi r^2}{4}} = \rho \frac{4l}{\pi r^2} \cdot 4 = 16 \rho \frac{l}{\pi r^2} = 16R. $$

Hence, the new resistance of the wire is $$16R$$, which means $$x = 16$$.